Since \(\eta = \mathcal {N}( {\bf 0}, {\bf I}_n )\) ,
a simple calculation shows that each component,
\(\widehat{S}^k\) , of \(\widehat{S}\)  can be computed independently as [1]}
\( \widehat{S}^k \in \arg \min _{U^k \in \mathcal {H}_k} \,\frac{1}{M} \sum _{i=1}^M \,\left(\frac{1}{2} \, U^k(z^i)^2- \log \partial _k U^k(z^i) \right),\) 
